qz1sol_6839sp10

# qz1sol_6839sp10 - 3 19.3 g/cm 3 = 19.3 g/cm 3 ×(1 kg/1000...

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TA: Tomoyuki Nakayama Tuesday, January 12 th , 2010 PHY 2048: Physic 1, Discussion Section 6839 Quiz 1 (Homework Set #1) Name: UFID: Formula sheets are not allowed. Calculators are allowed. Do not store equations in your calculator. You need to show all of your work for full credit. ________________________________________________________________________________ Gold has a density of 19.3 g/cm 3 , and the mass of a gold atom is 3.27 × 10 -25 kg. Assume the atoms are spherical and tightly packed. a) What is the volume of a gold atom? First we express the density in kg/m
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Unformatted text preview: 3 . 19.3 g/cm 3 = 19.3 g/cm 3 × (1 kg/1000 g) × (100 cm/1 m) 3 = 19.3 × 10 3 kg/m 3 Density is defined as mass divided by volume, thus the volume of a gold atom is V = m/ρ = (3.27 × 10-25 ) / (19.3 × 10 3 ) = 1.69 × 10-29 m 3 b) What is the distance between the centers of adjacent atoms? The distance between the centers of adjacent atoms is twice the radius of an atom, which is equal to the diameter. So we have V = (4π/3) × (d/2) 3 = πd 3 /6 d = ⇒ 3 √(6V/π) = 3 √(6 × 1.69 × 10-29 /π) = 3.18 × 10-10 m...
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